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3 years ago

How many mL of 6.00 M HCl are needed to prepare 1500 mL of 0.200 M HCl solution?

Chemistry

1 answer:

GaryK [48]3 years ago

8 0

Answer:

50mL

Explanation:

First find out how many moles will be needed in your final solution.

concentration (c) = Moles / Volume of solution (liters)

or rearrange the formula to find your Moles

Moles = concentration x volume of solution

convert your mL to L. 1500 mL is the same as 1.5 L

Moles = 0.200 moles/liter x 1.5 liter (the liters cancel each other out)

Moles = 0.3 moles

Now use the same formula to determine how much of the original solution you need to end up with what you want.

Rearrange the formula again to solve for the volume.

Volume of solution = Moles / concentration

Volume = 0.3 moles / 6.00 (moles/liter) here the moles cancel out

Volume = 0.05 L

convert your L into mL

Volume = 50 mL

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Ostrovityanka [42]

<h3>Answer:</h3>

The lowest boiling point is of n-Butane because it only experiences London Dispersion Forces between molecules.

<h3>Explanation:</h3>

Lets take start with the melting point of both compounds.

n-Butane = - 140 °C

Trimethylamine = - 117 °C

Intermolecular Forces in n-Butane:

As we know n-Butane is made up of Carbon and Hydrogen atoms only bonded via single covalent bonds. The electronegativity difference between C and C atoms is zero while, that between C and H atoms is 0.35 which is less than 0.4. Hence, the bonds in n-Butane are purely non polar in nature. Therefore, only London Dispersion Forces are found in n-Butane which are considered as the weakest intermolecular interactions.

Intermolecular Forces in Trimethylamine:

Trimethylamine (a tertiary amine) is made up of Nitrogen, Carbon and Hydrogen atoms bonded via single covalent bonds. The electronegativity difference between N and C atoms is 0.49 which is greater than 0.4. Hence, the C-N bond is polar in nature. Therefore, Dipole-Dipole interactions will be formed along with London Dispersion Forces which are stronger than Dispersion Forces. Therefore, due to Dipole-Dipole interactions Trimethylamine will have greater melting point than n-Butane.

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4 years ago

In at least 4 complete sentences, describe the similarities and differences between Avogadro's Law and Charles' Law.

viktelen [127]

answer

Avogadro's law states that, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles present. In other words, equal volumes of gases at the same pressure and temperature contain the same number of molecules - this is true regardless of their physical properties or chemical nature.

This number of molecules is

6.022

⋅

and is known as Avogadro's number,

Matematically, Avogadro's law can be written like this

, or, better yet,

Avogadro's law, as well as Boyle's law and Charles' law, are special cases of the ideal gas law,

. If temperature and pressure are kept constant, and knowing that

is of course constant, then

→

, which represents Avogadro's law.

The ideal gas law can also be written to incorporate

, since the number of moles are actually the number of molecules divided by Avogadro's number

⋅

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3 years ago

In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for

olga_2 [115]

<u>Answer:</u> The $E^o_{cell}\text{ and }K_{eq}$ of the reaction is 0.78 V and $2.44\times 10^{26}$ respectively.

<u>Explanation:</u>

For the given half reactions:

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.34V$

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To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

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$\Delta G^o=-nfE^o_{cell}$

and,

$\Delta G^o=-RT\ln K_{eq}$

Equating these two equations, we get:

$nfE^o_{cell}=RT\ln K_{eq}$

where,

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F = Faraday's constant = 96500 C

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R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = $25^oC=[273+25]=298K$

$K_{eq}$ = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

$2\times 96500\times 0.78=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=2.44\times 10^{26}$

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